Question: Suppose we want to be 99% confiodent that the sample proportion of adu

Question

Question: Suppose we want to be 99% confiodent that the sample proportion of adults who watch streaming programming to be within 2 percentage points of the true proportion of adults who watch streaming programming. How many additional college students must we sample based on the obtined sample proportions of 53%.

where n of original sample = 2343 adults surveyed

Im a little lost inthis question. More specifically the estimation error part.

Bascially to find n = (2*z/w)^2 pq then we subrtract this with the original num,ber of adults surveys to find the additoonal ones!

But for w or E whichever you prefer - its the estimation error- is 0.04 and not 0.02? Not sure how though. Why is it 2*2%- the logic is failing me here?

Explanation / Answer

a.

TRADITIONAL METHOD

given that,

possibile chances (x)=1241.79

sample size(n)=2343

success rate ( p )= x/n = 0.53

I.

sample proportion = 0.53

standard error = Sqrt ( (0.53*0.47) /2343) )

= 0.01

II.

margin of error = Z a/2 * (stanadard error)

where,

Za/2 = Z-table value

level of significance, = 0.01

from standard normal table, two tailed z /2 =2.576

margin of error = 2.576 * 0.01

= 0.027

III.

CI = [ p ± margin of error ]

confidence interval = [0.53 ± 0.027]

= [ 0.503 , 0.557]

-----------------------------------------------------------------------------------------------

DIRECT METHOD

given that,

possibile chances (x)=1241.79

sample size(n)=2343

success rate ( p )= x/n = 0.53

CI = confidence interval

confidence interval = [ 0.53 ± 2.576 * Sqrt ( (0.53*0.47) /2343) ) ]

= [0.53 - 2.576 * Sqrt ( (0.53*0.47) /2343) , 0.53 + 2.576 * Sqrt ( (0.53*0.47) /2343) ]

= [0.503 , 0.557]

-----------------------------------------------------------------------------------------------

interpretations:

1. We are 99% sure that the interval [ 0.503 , 0.557] contains the true population proportion

2. If a large number of samples are collected, and a confidence interval is created

for each sample, 99% of these intervals will contains the true population proportion

b.

true proportion is 2%

sample proportion is 53%

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.01 is = 2.576

Sample Proportion = 0.53

ME = 0.02

n = ( 2.576 / 0.02 )^2 * 0.53*0.47

= 4132.43 ~ 4133

c.

if you take estimate proportion is 0.04 then

sample size n=?

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.01 is = 2.576

Sample Proportion = 0.53

ME = 0.04

n = ( 2.576 / 0.04 )^2 * 0.53*0.47

= 1033.107 ~ 1034

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